If … then k is equal to:

Let $\Delta=\left|\begin{array}{ccc}a^2 & b^2 & c^2 \\ (a+\lambda)^2 & (b+\lambda)^2 & (c+\lambda)^2 \\ (a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2\end{array}\right|$Δ=​a2(a+λ)2(a−λ)2​b2(b+λ)2(b−λ)2​c2(c+λ)2(c−λ)2​​ $$\begin{aligned} &\text { Apply } \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3 \\ &\Delta=\left|\begin{array}{ccc} a^2 & b^2 & c^2 \\ (a+\lambda)^2-(a-\lambda)^2 & (b+\lambda)^2-(b-\lambda)^2 & (c+\lambda)^2-(c-\lambda)^2 \\ (a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2 \end{array}\right| \\ &=\left|\begin{array}{ccc} a^2 & b^2 & c^2 \\ 4 a \lambda & 4 b \lambda & 4 c \lambda \\ (a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2 \end{array}\right| \\ &\left(\because(x+y)^2-(x-y)^2=4 x y\right) \\ & \end{aligned}$$​ Apply R2​→R2​−R3​Δ=​a2(a+λ)2−(a−λ)2(a−λ)2​b2(b+λ)2−(b−λ)2(b−λ)2​c2(c+λ)2−(c−λ)2(c−λ)2​​=​a24aλ(a−λ)2​b24bλ(b−λ)2​c24cλ(c−λ)2​​(∵(x+y)2−(x−y)2=4xy)​ Taking out 4 common from $\mathrm{R}_2$R2​ $=4\left|\begin{array}{ccc}a^2 & b^2 & c^2 \\ a \lambda & b \lambda & c \lambda \\ a^2+\lambda^2-2 a \lambda & b^2+\lambda^2-2 b \lambda & c^2+\lambda^2-2 c \lambda\end{array}\right|$=4​a2aλa2+λ2−2aλ​b2bλb2+λ2−2bλ​c2cλc2+λ2−2cλ​​ $$\begin{aligned} &\text { Apply } \mathrm{R}_3 \rightarrow\left[\mathrm{R}_3-\left(\mathrm{R}_1-2 \mathrm{R}_2\right)\right] \\ &=4\left|\begin{array}{lll} a^2 & b^2 & c^2 \\ a \lambda & b \lambda & c \lambda \\ \lambda^2 & \lambda^2 & \lambda^2 \end{array}\right| \end{aligned}$$​ Apply R3​→[R3​−(R1​−2R2​)]=4​a2aλλ2​b2bλλ2​c2cλλ2​​​ Taking out $\lambda$λ common from $\mathrm{R}_2$R2​ and $\lambda^2$λ2 from $\mathrm{R}_3$R3​. $$=4 \lambda\left(\lambda^2\right)\left|\begin{array}{ccc} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{array}\right|$$=4λ(λ2)​a2a1​b2b1​c2c1​​ $$\begin{aligned} &=k \lambda\left|\begin{array}{ccc} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{array}\right| \\ &\Rightarrow k=4 \lambda^2 \end{aligned}$$​=kλ​a2a1​b2b1​c2c1​​⇒k=4λ2​